Question: Find $\dfrac{d}{dx}\left(\ln(\sqrt{x})\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac1{\sqrt{x}}$ (Choice B) B $\dfrac1{2x}$ (Choice C) C $\ln\left(\dfrac1{2\sqrt{x}}\right)$ (Choice D) D $\dfrac12\ln\left(\dfrac1{2\sqrt{x}}\right)$
Explanation: $\ln(\sqrt{x})$ is a composition of two, more basic, functions: $\sqrt{x}$ and $\ln(x)$. In other words, suppose $u(x)=\sqrt{x}$ and $v(x)=\ln(x)$, then $\ln(\sqrt{x})=v\Bigl(u(x)\Bigr)$, or $(v\circ u)(x)$. Therefore, $\dfrac{d}{dx}\left(\ln(\sqrt{x})\right)$ can be found using the chain rule : $\begin{aligned} \dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&=\dfrac{dv}{du}\cdot\dfrac{du}{dx} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x) \end{aligned}$ Finding $v'\Bigl(u(x)\Bigr)$ $v(x)=\ln(x)$, and therefore $v'(x)=\dfrac1x$. Now we plug $u(x)=\sqrt{x}$ into $v'$ : $\begin{aligned} v'\Bigl(u(x)\Bigr)&=v'\left(\sqrt{x}\right) \\\\ &={\dfrac1{\sqrt{x}}} \end{aligned}$ Finding $u'(x)$ $u(x)=\sqrt{x}$, and therefore $u'(x)={\dfrac1{2\sqrt{x}}}$. Putting things together $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\ln(\sqrt{x})\right) \\\\ &=\dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&&\gray{\text{Let }u(x)=\sqrt{x}\text{, }v(x)=\ln(x)} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x)&&\gray{\text{The chain rule}} \\\\ &={\dfrac1{\sqrt{x}}}\cdot {\dfrac1{2\sqrt{x}}} \\\\ &=\dfrac{1}{2x} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\ln(\sqrt{x})\right)=\dfrac{1}{2x}$.